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HOW HEATING EFFECT PRODUCED BY ELECTRIC CURRENT?
HOW HEATING EFFECT PRODUCED BY ELECTRIC CURRENT?
We know that a battery or a cell is a source of electrical energy. The chemical reaction within the cell generates the potential difference between its two terminals that sets the electrons in motion to flow the current through a resistor or a system of resistors connected to the battery. We have also seen, in Section 12.2, that to maintain the current, the source has to keep expending its energy. Where does this energy go? A part of the source energy in maintaining the current may be consumed into useful work (like in rotating the blades of an electric fan). Rest of the source energy may be expended in heat to raise the temperature of the gadget. We often observe this in our everyday life. For example, an electric fan becomes warm if used continuously for longer time etc. On the other hand, if the electric circuit is purely resistive, that is, a configuration of resistors only connected to a battery; the source energy continually gets dissipated entirely in the form of heat. This is known as heating effect of electric current. This effect is utilised in devices such as the electric heater, electric iron etc.
Figure 12.13: A steady current in a purely resistive electric circuit
Consider a current I flowing through a resistor of resistance R. Let the potential difference across it be V (Fig. 12.13). Let t be the time during which a charge Q flows across. The work is done in moving the charge Q through a potential difference V is VQ. Therefore, the source must supply energy equal to VQ in time t. Hence the power input to the circuit by the source is
P = V\(\frac{Q}{t}\) =VI ........ (12.19)
Or the energy supplied to the circuit by the source in time t is (P.t), that is, (V.I.t). What happens to this energy expended by the source? This energy gets dissipated in the resistor as heat. Thus for a steady current I, the amount of heat H produced in time t is
H = V\(\times\) I \(\times\) t ........ (12.20)
Applying Ohm’s law [Eq. (12.5)], we get
H = I2 Rt ........ (12.21)
This is known as Joule’s law of heating. The law implies that heat produced in a resistor is
- directly proportional to the square of current for a given resistance,
- directly proportional to resistance for a given current, and
- directly proportional to the time for which the current flows through the resistor. In practical situations, when an electric appliance is connected to a known voltage source, Eq. (12.21) is used after calculating the current through it, using the relation I = V/R.
Illustration 12.10:
An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. What are the current and the resistance in each case?
Sol:
From Eq. (12.19), we know that the power input is
P = V I
Thus the current I = P/V
(a) When heating is at the maximum rate,
I = 840 W/220 V = 3.82 A;
and the resistance of the electric iron is
R = V/I = 220 V/3.82 A = 57.60 \(\Omega\).
(b) When heating is at the minimum rate,
I = 360 W/220 V = 1.64 A;
and the resistance of the electric iron is
R = V/I = 220 V/1.64 A = 134.15 \(\Omega\).
Illustration 12.11:
100 J of heat is produced each second in a 4 \(\Omega\) resistance. Find the potential difference across the resistor.
Sol:
H = 100 J, R = 4 \(\Omega\), t = 1 s, V = ?
From Eq. (12.21) we have the current through the resistor as
I = \(\sqrt{(H/Rt)}\)
= \(\sqrt{100J/(4\Omega\times1s)}\)
= 5 A
Thus the potential difference across the resistor, V [from Eq. (12.5)] is
V = I R
= 5 A \(\times\) 4 \(\Omega\)
= 20 V.
Questions
1. Why does the cord of an electric heater not glow while the heating element does?
2. Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V.
3. An electric iron of resistance 20 \(\Omega \) takes a current of 5 A. Calculate the heat developed in 30 s.
Source: This topic is taken from NCERT TEXTBOOK
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PRACTICAL APPLICATIONS
PRACTICAL APPLICATIONS
The generation of heat in a conductor is an inevitable consequence of electric current. In many cases, it is undesirable as it converts useful electrical energy into heat. In electric circuits, the unavoidable heating can increase the temperature of the components and alter their properties. However, heating effect of electric current has many useful applications. The electric iron, electric toaster, electric oven, electric kettle and electric heater are some of the familiar devices based on Joule’s heating.
The electric heating is also used to produce light, as in an electric bulb. Here, the filament must retain as much of the heat generated as is possible, so that it gets very hot and emits light. It must not melt at such high temperature. A strong metal with a high melting point such as tungsten (melting point 3380°C) is used for making bulb filaments. The filament should be thermally isolated as much as possible, using insulating support, etc. The bulbs are usually filled with chemically inactive nitrogen and argon gases to prolong the life of the filament. Most of the power consumed by the filament appears as heat, but a small part of it is in the form of light radiated.
Another common application of Joule’s heating is the fuse used in electric circuits. It protects circuits and appliances by stopping the flow of any unduly high electric current. The fuse is placed in series with the device. It consists of a piece of wire made of a metal or an alloy of an appropriate melting point, for example, aluminium, copper, iron, lead, etc. If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit.
The fuse wire is usually encased in a cartridge of porcelain or similar material with metal ends. The fuses used for domestic purposes are rated as 1 A, 2 A, 3 A, 5 A, 10 A, etc. For an electric iron which consumes 1 kW electric power when operated at 220 V, a current of (1000/220) A, that is, 4.54 A will flow in the circuit. In this case, a 5 A fuse must be used.
Source: This topic is taken from NCERT TEXTBOOK